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# hw2_soln - Physics 313 Problem Set 2 Due Friday by the end...

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Physics 313 Problem Set 2 Due Friday September 22, 2006 by the end of class 1. Schroeder Problem 1.36 A. Use V γ P =constant to get (1 L ) γ (1 atm ) = V γ (7 atm ), so V = (1 / 7) (1 ) . Here γ = 1 . 4 for air, so V = 0 . 249 L. B. To get the work, do W = R P ( V ) dV . Now V γ P =constant = 1 to start with (in units of literxs and atmospheres), so P = V - γ . W = - Z V f V i dV V - γ = 1 γ - 1 V - γ +1 f - V - γ +1 i = 1 0 . 4 ( 0 . 249 - 0 . 4 - 1 - 0 . 4 ) = 1 . 86 So W = 1 . 86 L · atm, which is 188 J. C. I assume the initial temperature is 300 K. The volume decreased from 1 L to 0.249 L, but the pressure increased by a factor of 7. Using the ideal gas law, the final temperature should scale with P and V, and so T = (300 K )(0 . 249)(7) = 523 K. 1

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2. Schroeder Problem 1.48 The incident power is 1000 W/m 2 , but 90% is reflected, so the absorbed power is 100 W/m 2 . The snow pack is 2 m thick. Imagine a 1 m 2 patch of the snow. We’re told it’s 50% air, so the total volume of ice to be melted is (2)(0 . 5) = 1 m 2 = 10 6 cm 2 . Ice has a densityof 0.9, so this is 0 . 9 × 10 6 = 9 × 10 5 grams of ice to melt. Ice has a latent heat of 333 J/g, so melting all of the ice would require (333)(9 × 10 5 ) = 3 × 10 8 J. The ice pack would therefore last for (3 × 10 8 J ) / 100 W = 3 × 10 6 seconds. In the spring (around equinox),
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hw2_soln - Physics 313 Problem Set 2 Due Friday by the end...

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